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ak2121
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What is the sum of all 4-digit numbers that can be formed 01 Oct 2021, 05:00
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Bunuel
dimitri92
I am sorry I don't have the OA. But I think it is solvable without the OA

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.

Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - \(\frac{4^4}{4}=4^3=64\) times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.

So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the \(n\) digits (repetition being allowed) is: \(n^{n-1}\)*(sum of the digits)*(111…..n times).

Hope it helps.
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Hi Bunuel, can you explain the logical way like you did for when the repetition is allowed for when the repetition is not allowed?
I am unable to figure it out. Just curious
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What is the sum of all 4-digit numbers that can be formed 01 Oct 2021, 14:34
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Hi ak2121,

You can calculate that total in the same general way - it's just that there wouldn't be as many overall numbers to consider. Here's how:

When forming 4-digit numbers with the digits 1, 2, 3 and 4 - but with NO repetitions - we have (4)(3)(2)(1) = 24 possible numbers. Since each digit can be in any of the 4 'spots', this means that we'll have...

6 numbers that begin with a 4
6 numbers that begin with a 3
6 numbers that begin with a 2
6 numbers that begin with a 1

That will hold true for each of the digits in the 4-digit number, so the overall sum of all those thousands, hundreds, tens and units digits would be...

6(4+3+2+1)(1000) + 6(4+3+2+1)(100) + 6(4+3+2+1)(10) + 6(4+3+2+1)(1)

You can actually simplify this in advance if you realize that 4+3+2+1 = 10...

6(10)(1000) + 6(10)(100) + 6(10)(10) + 6(10)(1) = 66,660

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What is the sum of all 4-digit numbers that can be formed Updated on: 25 Jun 2026, 08:03
Originally posted by ozzaayy on 18 Jun 2024, 14:52.
Last edited by Bunuel on 25 Jun 2026, 08:03, edited 1 time in total.
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What is the sum of all 4-digit numbers that can be formed 07 Jul 2026, 23:55
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Hello! How do we solve this question if 0 were one of the digits?
Bunuel


As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.

Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - \(\frac{4^4}{4}=4^3=64\) times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.

So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(sum of the digits)*(111.....n times).

2. Sum of all the numbers which can be formed by using the \(n\) digits (repetition being allowed) is: \(n^{n-1}\)*(sum of the digits)*(111.....n times).

Hope it helps.
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